(1st) if there is NO Wait times at city B or City C, the entire trip would take:
25 + 35 + 60 = 120 minutes
(2nd) on top of the 120 minutes of Travel Time, we want to MINIMIZE the Wait Time we spend waiting for the Next Departure in 2 Places:
-1- when we arrive in City B - waiting to Depart City B
and
-2- when we arrive in City C - waiting to Depart City C
Method: usually follow a few Departure Times starting from A and travel through City-to-City, Minimizing the Wait Time for the Next Departure in each Case ------> often, a Pattern will develop
Case 1:
Leave at 5:20 from A
Arrive at 5:45 at B
the Earliest Bus that will leave is at 6:00 ------ (wait 15 minutes)
Arrive at 6:35 at C
the Earliest Bus that will leave is at 6:40 ------ (wait 5 minutes)
Total Travel Times + Total Wait Time = 120 + (15 + 5) = 140 minutes
Eliminate -D- and -E-
Also Eliminate Answer -A- 120 minutes
just looking at the Possible Departures from A (and the fact it takes 25 minutes to get from A-to-B) the MIN wait time in City B will be 5 minutes. Thus, it could never possibly take 120 minutes
Can we get there in less time than 140 minutes?
case 2:
leave A at 5:40
Arrive at B at 6:05
Earliest Bus leaves at 6:30 ------ (25 minute wait time)
Arrive at C at 7:05
Earliest Bus leaves at 7:20 ------ (15 minute wait time)
Total Travel Time + Wait Time = 120 + (25 + 15) = 160 minutes
Case 3:
leave A at 6:00
Arrive at B at 6:25
Earliest Departure is at 6:30 ------ (5 minute wait time)
Arrive at C at 7:05
Earliest Departure is at 7:20 ----- (15 minute wait time)
120 minutes + 20 minutes = 140 minutes
If you keep alternating Departure Times from A every 20 minutes and then MINIMIZE the Time spent waiting for the next departure at Each City,
the Minimum Possible Journey Times will alternate between 140 minutes and 160 minutes
thus, Answer is -C-
140 minutes is the MIN